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Contact [7]
1 year ago
13

You drag a crate across a floor with a rope. The force applied is 750 N and the angle of the rope is 25.0° above the horizontalH

ow much work do you do on the crate when moving it 26.0 m?If you complete the job in 6.00 s, what power is developed?
Physics
1 answer:
frez [133]1 year ago
5 0

a.

The work done by a constant force along a rectilinear motion when the force and the displacement vector are not colinear is given by:

W=F\cos\theta\cdot d

where F is the magnitude of the force, theta is the angle between them and d is the distance.

The problen gives the following data:

The magnitude of the force 750 N.

The angle between the force and the displacement which is 25°

The distance, 26 m.

Plugging this in the formula we have:

\begin{gathered} W=\left(750\right)\left(\cos25\right)\left(26\right) \\ W=17673 \end{gathered}

Therefore the work done is 17673 J.

b)

The power is given by:

P=\frac{W}{t}

the problem states that the time it takes is 6 s. Then:

\begin{gathered} P=\frac{17673}{6} \\ P=2945.5 \end{gathered}

Therefore the power is 2945.5 W

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Answer:

Current, I = 0.0011 A

Explanation:

It is given that,

Diameter of rod, d = 2.56 cm

Radius of rod, r = 1.28 cm = 0.0128 m

The resistivity of the pure silicon, \rho=2300\ \Omega-m

Length of rod, l = 20 cm = 0.2 m

Voltage, V=1\times 10^3\ V

The resistivity of the rod is given by :

R=\rho\dfrac{L}{A}

R=2300\ \Omega-m\dfrac{0.2\ m}{\pi (0.0128\ m)^2}

R = 893692.30 ohms

Current flowing in the rod is calculated using Ohm's law as :

V = I R

I=\dfrac{V}{R}

I=\dfrac{10^3\ V}{893692.30\ \Omega}

I = 0.0011 A

So, the current flowing in the rod is 0.0011 A. Hence, this is the required solution.

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3 years ago
Perfect symmetrical cone type volcanoes would have what type of drainage pattern
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Perfect symmetrical cone type volcanoes would have radial drainage pattern.

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A ring, cylinder, solid sphere, and hollow sphere are all released from rest from the same height on an inclined surface, at the
crimeas [40]

Answer:

Explanation:

  • The expression for acceleration of the rolling body on an inclined plane is given as a = gsinФ/1 + k²/R²
  • where Ф is the angle of inclination, R is the radius, k is the radius of gyration.
  • The potential energy of the system is given as ; PE = mgh
  • The potential energy will be constant for ring, cylinder, solid sphere, and hollow sphere.
  • The total kinetic energy of the rolling body is ; KE = mv²/2 + Iw²/2
  • Hence, the total kinetic energy of the ring, cylinder, solid sphere and hollow sphere will be constant.

2. The moment of inertia of the ring is given as ;

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The moment of inertia of the ring is maximum and therefore reaches the bottom last.

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3 years ago
Physics Help ASAP Please!
balu736 [363]

Answer:

23N

Explanation:

5 0
3 years ago
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Normal atmospheric pressure is 1.013 105 Pa. The approach of a storm causes the height of a mercury barometer to drop by 27.1 mm
Burka [1]

Answer:

The atmospheric pressure is 0.97622\times10^{5}\ Pa.

Explanation:

Given that,

Atmospheric pressure P_{atm}= 1.013\times10^{5}\ Pa    

drop height h'= 27.1 mm

Density of mercury \rho= 13.59 g/cm^3

We need to calculate the height

Using formula of pressure

p = \rho g h

Put the value into the formula

1.013\times10^{5}=13.59\times10^{3}\times9.8\times h

h =\dfrac{1.013\times10^{5}}{13.59\times10^{3}\times9.8}

h=0.76\ m

We need to calculate the new height

h''=h - h'

h''=0.76-27.1\times10^{-3}

h''=0.76-0.027

h''=0.733\ m

We need to calculate the atmospheric pressure

Using formula of atmospheric pressure

P=\rho g h

Put the value into the formula

P= 13.59\times10^{3}\times9.8\times0.733

P=0.97622\times10^{5}\ Pa

Hence, The atmospheric pressure is 0.97622\times10^{5}\ Pa.

7 0
3 years ago
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